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Matrices and Determinants question

2022 · 29 Jun · Shift 1 · Q27
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Matrices and Determinants question

2022 · 29 Jun · Shift 1 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=[aij]A = [{a_{ij}}]A=[aij​] be a square matrix of order 3 such that aij=2j−i{a_{ij}} = {2^{j - i}}aij​=2j−i, for all i, j = 1, 2, 3. Then, the matrix A2 + A3 + ...... + A10 is equal to :
  1. A
    (310−32)A\left( {{{{3^{10}} - 3} \over 2}} \right)A(2310−3​)A
  2. B
    (310−12)A\left( {{{{3^{10}} - 1} \over 2}} \right)A(2310−1​)A
  3. C
    (310+12)A\left( {{{{3^{10}} + 1} \over 2}} \right)A(2310+1​)A
  4. D
    (310+32)A\left( {{{{3^{10}} + 3} \over 2}} \right)A(2310+3​)A
View written solutionFree

Correct answer: A

  1. Write the matrix explicitly

Given aij=2j−i, i,j=1,2,3a_{ij}=2^{j-i}, \, i,j=1,2,3aij​=2j−i,i,j=1,2,3 so

2^{1-1} & 2^{2-1} & 2^{3-1}\\ 2^{1-2} & 2^{2-2} & 2^{3-2}\\ 2^{1-3} & 2^{2-3} & 2^{3-3} \end{pmatrix} =\begin{pmatrix} 1 & 2 & 4\\ \tfrac12 & 1 & 2\\ \tfrac14 & \tfrac12 & 1 \end{pmatrix}.$$ 2. **Observe the structure of entries** The $(i,j)$-th entry is $$a_{ij}=2^{j-i}=2^{-i}\cdot 2^j.$$ So we can write $$A=uv^T$$ where $$u=\begin{pmatrix}2^{-1}\\2^{-2}\\2^{-3}\end{pmatrix}, \qquad v=\begin{pmatrix}2^1\\2^2\\2^3\end{pmatrix}.$$ Indeed, $$(uv^T)_{ij}=u_i v_j=2^{-i}2^j=2^{j-i}=a_{ij}.$$ Thus $A$ is a rank-one matrix. 3. **Use the rank-one power property** For a matrix of the form $uv^T$, $$(uv^T)^2=u(v^Tu)v^T=(v^Tu)(uv^T).$$ So first compute $$v^Tu=2^1\cdot 2^{-1}+2^2\cdot 2^{-2}+2^3\cdot 2^{-3}=1+1+1=3.$$ Hence $$A^2=3A.$$ Now repeatedly, $$A^3=A\cdot A^2=A\cdot 3A=3A^2=3(3A)=3^2A,$$ and in general, $$A^n=3^{n-1}A \quad (n\ge 1).$$ 4. **Compute the required sum** We need $$A^2+A^3+\cdots +A^{10}.$$ Using $A^n=3^{n-1}A$, $$A^2+A^3+\cdots +A^{10} =(3A+3^2A+\cdots +3^9A).$$ Factor out $A$: $$= (3+3^2+\cdots +3^9)A.$$ This is a geometric series with first term $3$, common ratio $3$, and $8$ terms: $$3+3^2+\cdots +3^9=\frac{3(3^8-1)}{3-1}=\frac{3^9-3}{2}\cdot 3?$$ Let us use the standard formula carefully: $$\sum_{k=1}^{9}3^k=\frac{3^{10}-3}{3-1}=\frac{3^{10}-3}{2}.$$ But we need from $k=1$ to $9$, exactly the same sum. Therefore $$3+3^2+\cdots +3^9=\frac{3^{10}-3}{2}.$$ So, $$A^2+A^3+\cdots +A^{10}=\left(\frac{3^{10}-3}{2}\right)A.$$ 5. **Match with the options** This matches **Option A**: $$\left(\frac{3^{10}-3}{2}\right)A.$$ Therefore, the correct answer is **A**.
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